Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/71

or than

or than

where ${\bar {\beta }}$ denotes the greatest of the numbers $|\beta _{1}|,|\beta _{2}|,\ldots |\beta _{n}|$.

It thus appears that the modulus of

$\sum \limits _{r=p-1}^{r=np+p-1}c_{r}{\beta _{m}}^{r}\left\{{\frac {\beta _{m}}{r+1}}+{\frac {{\beta _{m}}^{2}}{(r+1)(r+2)}}+\ldots \right\}$

is less than a number of the form $PQ^{P}/(p-1)!$, where $P$ and $Q$ are independent of $p$ and of $m$.

We have now

(A + T A) - KPA + T #<*> (ft,)

T

m=l m=l

where $K_{p}A$ is not a multiple of $p$, the second term is an integer divisible by $p$, and $L$ is less than $nPQ^{P}/(p-1)!$. The prime $p$ may be chosen so large that $nPQ^{P}/(p-1)!$ is numerically less than unity. Since $\textstyle K_{p}(A+\sum \limits _{m=1}^{m=n}e^{\beta _{m}})$ is expressed as the sum of an integer which does not vanish and of a number numerically less than unity, it is impossible that it can vanish. Having now shewn that no such equation as

$A+e^{\beta _{1}}+e^{\beta _{2}}+\ldots +e^{\beta _{m}}=0$

can subsist, we see that $\pi i$ cannot be a root of an algebraic equation with integral coefficients, and thus that $\pi$ is transcendental.

It has thus been proved that $\pi$ is a transcendental number, and hence, taking into account the theorem proved on page 50, the im- possibility of "squaring the circle" has been effectively established.

CAMBRIDGE: PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS